Leetcode: reverse list

Method One: iterative

The method defines the head node and tail node two variables, by nextTemp record pointer, the pointer sequence 1-> 2-> 3, a second step of modifying reference <-2 <-3 iteratively.

. 1  public ListNode reverseList (ListNode head) {
 2      // tail node 
. 3      ListNode PREV = null ;
 . 4      // head node 
. 5      ListNode Curr = head;
 . 6      the while (! Curr = null ) {
 . 7          // record the tail current node 
8          nextTemp = ListNode curr.next;
 . 9          // modify the current node is tail node null, (l-> null, 2-> l-> null, 3-> 2-> l-> null) 
10          curr.next = PREV;
 11          // record the next node in the end node 
12 is          PREV = Curr;
 13 is          //The next section nod recording node (original node is tail node) 
14          Curr = nextTemp;
 15      }
 16      return PREV;
 . 17 }

Complexity Analysis

• Time complexity: O (n-) . Suppose  n- n-is the length of the list, the time complexity is  O (n-) .

• Complexity Space: O (. 1) .